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Discrete MathematicsGraphs
H. Turgut Uyar Aysegul Gencata Yayımlı Emre Harmancı
2001-2016
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c© 2001-2016 T. Uyar, A. Yayımlı, E. Harmancı
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Topics
1 GraphsIntroductionWalksTraversable GraphsPlanar Graphs
2 Graph ProblemsConnectivityGraph ColoringShortest PathTSPSearching Graphs
Topics
1 GraphsIntroductionWalksTraversable GraphsPlanar Graphs
2 Graph ProblemsConnectivityGraph ColoringShortest PathTSPSearching Graphs
Graphs
Definition
graph: G = (V ,E )
V : node (or vertex) set
E ⊆ V × V : edge set
e = (v1, v2) ∈ E :
v1 and v2 are endnodes of e
e is incident to v1 and v2
v1 and v2 are adjacent
Graphs
Definition
graph: G = (V ,E )
V : node (or vertex) set
E ⊆ V × V : edge set
e = (v1, v2) ∈ E :
v1 and v2 are endnodes of e
e is incident to v1 and v2
v1 and v2 are adjacent
Graph Example
V = {a, b, c, d , e, f }E = {(a, b), (a, c),
(a, d), (a, e),(a, f ), (b, c),(d , e), (e, f )}
Directed Graphs
Definition
directed graph (digraph): D = (V ,A)
V : node set
A ⊆ V × V : arc set
a = (v1, v2) ∈ A:
v1: origin node of a
v2: terminating node of a
Directed Graph Example
Weighted Graphs
weighted graph: labels assigned to edges
weight
length, distance
cost, delay
probability
. . .
Multigraphs
parallel edges: edges between same node pair
loop: edge starting and ending in same node
plain graph: no loops, no parallel edges
multigraph: a graph which is not plain
Multigraph Example
parallel edges: (a, b)
loop: (e, e)
Subgraph
Definition
G ′ = (V ′,E ′) is a subgraph of G = (V ,E ):V ′ ⊆ V ∧ E ′ ⊆ E ∧ ∀(v1, v2) ∈ E ′ [v1, v2 ∈ V ′]
Incidence Matrix
rows: nodes, columns: edges
1 if edge incident on node, 0 otherwise
example
e1 e2 e3 e4 e5 e6 e7 e8
v1 1 1 1 0 1 0 0 0v2 1 0 0 1 0 0 0 0v3 0 0 1 1 0 0 1 1v4 0 0 0 0 1 1 0 1v5 0 1 0 0 0 1 1 0
Adjacency Matrix
rows: nodes, columns: nodes
1 if nodes are adjacent, 0 otherwise
example
v1 v2 v3 v4 v5
v1 0 1 1 1 1v2 1 0 1 0 0v3 1 1 0 1 1v4 1 0 1 0 1v5 1 0 1 1 0
Adjacency Matrix
multigraph: number of edges between nodes
example
a b c d
a 0 0 0 1b 2 1 1 0c 0 0 0 0d 0 1 1 0
Adjacency Matrix
weighted graph: weight of edge
Degree
degree of node: number of incident edges
Theorem
di : degree of vi
|E | =∑
i di
2
Degree
degree of node: number of incident edges
Theorem
di : degree of vi
|E | =∑
i di
2
Degree Example
da = 5db = 2dc = 2dd = 2de = 3df = 2
16|E | = 8
Multigraph Degree Example
da = 6db = 3dc = 2dd = 2de = 5df = 2
20|E | = 10
Degree in Directed Graphs
in-degree: dvi
out-degree: dvo
node with in-degree 0: source
node with out-degree 0: sink∑v∈V dv
i =∑
v∈V dvo = |A|
Degree in Directed Graphs
in-degree: dvi
out-degree: dvo
node with in-degree 0: source
node with out-degree 0: sink∑v∈V dv
i =∑
v∈V dvo = |A|
Degree in Directed Graphs
in-degree: dvi
out-degree: dvo
node with in-degree 0: source
node with out-degree 0: sink∑v∈V dv
i =∑
v∈V dvo = |A|
Degree
Theorem
In an undirected graph, there is an even number of nodeswhich have an odd degree.
Proof.
ti : number of nodes of degree i2|E | =
∑i di = 1t1 + 2t2 + 3t3 + 4t4 + 5t5 + . . .
2|E | − 2t2 − 4t4 − · · · = t1 + t3 + t5 + · · ·+ 2t3 + 4t5 + . . .2|E | − 2t2 − 4t4 − · · · − 2t3 − 4t5 − · · · = t1 + t3 + t5 + . . .
left-hand side even ⇒ right-hand side even
Degree
Theorem
In an undirected graph, there is an even number of nodeswhich have an odd degree.
Proof.
ti : number of nodes of degree i2|E | =
∑i di = 1t1 + 2t2 + 3t3 + 4t4 + 5t5 + . . .
2|E | − 2t2 − 4t4 − · · · = t1 + t3 + t5 + · · ·+ 2t3 + 4t5 + . . .2|E | − 2t2 − 4t4 − · · · − 2t3 − 4t5 − · · · = t1 + t3 + t5 + . . .
left-hand side even ⇒ right-hand side even
Degree
Theorem
In an undirected graph, there is an even number of nodeswhich have an odd degree.
Proof.
ti : number of nodes of degree i2|E | =
∑i di = 1t1 + 2t2 + 3t3 + 4t4 + 5t5 + . . .
2|E | − 2t2 − 4t4 − · · · = t1 + t3 + t5 + · · ·+ 2t3 + 4t5 + . . .2|E | − 2t2 − 4t4 − · · · − 2t3 − 4t5 − · · · = t1 + t3 + t5 + . . .
left-hand side even ⇒ right-hand side even
Degree
Theorem
In an undirected graph, there is an even number of nodeswhich have an odd degree.
Proof.
ti : number of nodes of degree i2|E | =
∑i di = 1t1 + 2t2 + 3t3 + 4t4 + 5t5 + . . .
2|E | − 2t2 − 4t4 − · · · = t1 + t3 + t5 + · · ·+ 2t3 + 4t5 + . . .2|E | − 2t2 − 4t4 − · · · − 2t3 − 4t5 − · · · = t1 + t3 + t5 + . . .
left-hand side even ⇒ right-hand side even
Degree
Theorem
In an undirected graph, there is an even number of nodeswhich have an odd degree.
Proof.
ti : number of nodes of degree i2|E | =
∑i di = 1t1 + 2t2 + 3t3 + 4t4 + 5t5 + . . .
2|E | − 2t2 − 4t4 − · · · = t1 + t3 + t5 + · · ·+ 2t3 + 4t5 + . . .2|E | − 2t2 − 4t4 − · · · − 2t3 − 4t5 − · · · = t1 + t3 + t5 + . . .
left-hand side even ⇒ right-hand side even
Degree
Theorem
In an undirected graph, there is an even number of nodeswhich have an odd degree.
Proof.
ti : number of nodes of degree i2|E | =
∑i di = 1t1 + 2t2 + 3t3 + 4t4 + 5t5 + . . .
2|E | − 2t2 − 4t4 − · · · = t1 + t3 + t5 + · · ·+ 2t3 + 4t5 + . . .2|E | − 2t2 − 4t4 − · · · − 2t3 − 4t5 − · · · = t1 + t3 + t5 + . . .
left-hand side even ⇒ right-hand side even
Isomorphism
Definition
G = (V ,E ) and G ? = (V ?,E ?) are isomorphic:∃f : V → V ? [(u, v) ∈ E ⇒ (f (u), f (v)) ∈ E ?] ∧ f is bijective
G and G ? can be drawn the same way
Isomorphism
Definition
G = (V ,E ) and G ? = (V ?,E ?) are isomorphic:∃f : V → V ? [(u, v) ∈ E ⇒ (f (u), f (v)) ∈ E ?] ∧ f is bijective
G and G ? can be drawn the same way
Isomorphism Example
f = (a 7→ d , b 7→ e, c 7→ b, d 7→ c, e 7→ a)
Isomorphism Example
f = (a 7→ d , b 7→ e, c 7→ b, d 7→ c, e 7→ a)
Petersen Graph
f = (a 7→ q, b 7→ v , c 7→ u, d 7→ y , e 7→ r ,f 7→ w , g 7→ x , h 7→ t, i 7→ z , j 7→ s)
Homeomorphism
Definition
G = (V ,E ) and G ? = (V ?,E ?) are homeomorphic:
G and G ? isomorphic, except that
some edges in E ? are divided with additional nodes
Homeomorphism Example
Regular Graphs
regular graph: all nodes have the same degree
n-regular: all nodes have degree n
examples
Completely Connected Graphs
G = (V ,E ) is completely connected:∀v1, v2 ∈ V (v1, v2) ∈ E
every pair of nodes are adjacent
Kn: completely connected graph with n nodes
Completely Connected Graph Examples
K4 K5
Bipartite Graphs
G = (V ,E ) is bipartite: V1 ∪ V2 = V , V1 ∩ V2 = ∅∀(v1, v2) ∈ E [v1 ∈ V1 ∧ v2 ∈ V2]
example
complete bipartite: ∀v1 ∈ V1 ∀v2 ∈ V2 (v1, v2) ∈ E
Km,n: |V1| = m, |V2| = n
Complete Bipartite Graph Examples
K2,3 K3,3
Topics
1 GraphsIntroductionWalksTraversable GraphsPlanar Graphs
2 Graph ProblemsConnectivityGraph ColoringShortest PathTSPSearching Graphs
Walk
walk: sequence of nodes and edgesfrom a starting node (v0) to an ending node (vn)
v0e1−→ v1
e2−→ v2e3−→ v3 −→ · · · −→ vn−1
en−→ vn
where ei = (vi−1, vi )
no need to write the edges if not weighted
length: number of edges
v0 = vn: closed
Walk
walk: sequence of nodes and edgesfrom a starting node (v0) to an ending node (vn)
v0e1−→ v1
e2−→ v2e3−→ v3 −→ · · · −→ vn−1
en−→ vn
where ei = (vi−1, vi )
no need to write the edges if not weighted
length: number of edges
v0 = vn: closed
Walk Example
c(c,b)−−−→ b
(b,a)−−−→ a(a,d)−−−→ d
(d ,e)−−−→ e(e,f )−−−→ f
(f ,a)−−−→ a(a,b)−−−→ b
c → b → a→ d → e→ f → a→ b
length: 7
Trail
trail: edges not repeated
circuit: closed trail
spanning trail: covers all edges
Trail Example
c(c,b)−−−→ b
(b,a)−−−→ a(a,e)−−−→ e
(e,d)−−−→ d(d ,a)−−−→ a
(a,f )−−−→ f
c → a→ e → d → a→ f
Path
path: nodes not repeated
cycle: closed path
spanning path: visits all nodes
Path Example
c(c,b)−−−→ b
(b,a)−−−→ a(a,d)−−−→ d
(d ,e)−−−→ e(e,f )−−−→ f
c → b → a→ d → e → f
Connected Graphs
connected: a path between every pair of nodes
a disconnected graph can be dividedinto connected components
Connected Components Example
disconnected:no path between a and c
connected components:a, d , eb, cf
Distance
distance between vi and vj :length of shortest path between vi and vj
diameter of graph: largest distance in graph
Distance Example
distance between a and e: 2
diameter: 3
Cut-Points
G − v : delete v and all its incident edges from G
v is a cut-point for G :G is connected but G − v is not
Cut-Point Example
G G − d
Directed Walks
ignoring directions on arcs: semi-walk, semi-trail, semi-path
if between every pair of nodes there is:
a semi-path: weakly connected
a path from one to the other: unilaterally connected
a path: strongly connected
Directed Graph Examples
weakly unilaterally strongly
Topics
1 GraphsIntroductionWalksTraversable GraphsPlanar Graphs
2 Graph ProblemsConnectivityGraph ColoringShortest PathTSPSearching Graphs
Bridges of Konigsberg
cross each bridge exactly onceand return to the starting point
Graphs
Traversable Graphs
G is traversable: G contains a spanning trail
a node with an odd degree must be either the starting nodeor the ending node of the trail
all nodes except the starting node and the ending nodemust have even degrees
Bridges of Konigsberg
all nodes have odd degrees: not traversable
Traversable Graph Example
a, b, c: even
d , e: odd
start from d , end at e:d → b → a→ c → e→ d → c → b → e
Euler Graphs
Euler graph: contains closed spanning trail
G is an Euler graph ⇔ all nodes in G have even degrees
Euler Graph Examples
Euler not Euler
Hamilton Graphs
Hamilton graph: contains a closed spanning path
Hamilton Graph Examples
Hamilton not Hamilton
Topics
1 GraphsIntroductionWalksTraversable GraphsPlanar Graphs
2 Graph ProblemsConnectivityGraph ColoringShortest PathTSPSearching Graphs
Planar Graphs
Definition
G is planar:G can be drawn on a plane without intersecting its edges
a map of G : a planar drawing of G
Planar Graph Example
Regions
map divides plane into regions
degree of region: length of closed trail that surrounds region
Theorem
dri : degree of region ri
|E | =∑
i dri
2
Regions
map divides plane into regions
degree of region: length of closed trail that surrounds region
Theorem
dri : degree of region ri
|E | =∑
i dri
2
Region Example
dr1 = 3dr2 = 3dr3 = 5dr4 = 4dr5 = 3
= 18|E | = 9
Euler’s Formula
Theorem (Euler’s Formula)
G = (V ,E ): planar, connected graphR: set of regions in a map of G
|V | − |E |+ |R| = 2
Euler’s Formula Example
|V | = 6, |E | = 9, |R| = 5
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3|E | ≤ 3|V | − 6
Proof.
sum of region degrees: 2|E |degree of a region ≥ 3⇒ 2|E | ≥ 3|R| ⇒ |R| ≤ 2
3 |E ||V | − |E |+ |R| = 2⇒ |V | − |E |+ 2
3 |E | ≥ 2 ⇒ |V | − 13 |E | ≥ 2
⇒ 3|V | − |E | ≥ 6 ⇒ |E | ≤ 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3|E | ≤ 3|V | − 6
Proof.
sum of region degrees: 2|E |degree of a region ≥ 3⇒ 2|E | ≥ 3|R| ⇒ |R| ≤ 2
3 |E ||V | − |E |+ |R| = 2⇒ |V | − |E |+ 2
3 |E | ≥ 2 ⇒ |V | − 13 |E | ≥ 2
⇒ 3|V | − |E | ≥ 6 ⇒ |E | ≤ 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3|E | ≤ 3|V | − 6
Proof.
sum of region degrees: 2|E |degree of a region ≥ 3⇒ 2|E | ≥ 3|R| ⇒ |R| ≤ 2
3 |E ||V | − |E |+ |R| = 2⇒ |V | − |E |+ 2
3 |E | ≥ 2 ⇒ |V | − 13 |E | ≥ 2
⇒ 3|V | − |E | ≥ 6 ⇒ |E | ≤ 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3|E | ≤ 3|V | − 6
Proof.
sum of region degrees: 2|E |degree of a region ≥ 3⇒ 2|E | ≥ 3|R| ⇒ |R| ≤ 2
3 |E ||V | − |E |+ |R| = 2⇒ |V | − |E |+ 2
3 |E | ≥ 2 ⇒ |V | − 13 |E | ≥ 2
⇒ 3|V | − |E | ≥ 6 ⇒ |E | ≤ 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3|E | ≤ 3|V | − 6
Proof.
sum of region degrees: 2|E |degree of a region ≥ 3⇒ 2|E | ≥ 3|R| ⇒ |R| ≤ 2
3 |E ||V | − |E |+ |R| = 2⇒ |V | − |E |+ 2
3 |E | ≥ 2 ⇒ |V | − 13 |E | ≥ 2
⇒ 3|V | − |E | ≥ 6 ⇒ |E | ≤ 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3|E | ≤ 3|V | − 6
Proof.
sum of region degrees: 2|E |degree of a region ≥ 3⇒ 2|E | ≥ 3|R| ⇒ |R| ≤ 2
3 |E ||V | − |E |+ |R| = 2⇒ |V | − |E |+ 2
3 |E | ≥ 2 ⇒ |V | − 13 |E | ≥ 2
⇒ 3|V | − |E | ≥ 6 ⇒ |E | ≤ 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3|E | ≤ 3|V | − 6
Proof.
sum of region degrees: 2|E |degree of a region ≥ 3⇒ 2|E | ≥ 3|R| ⇒ |R| ≤ 2
3 |E ||V | − |E |+ |R| = 2⇒ |V | − |E |+ 2
3 |E | ≥ 2 ⇒ |V | − 13 |E | ≥ 2
⇒ 3|V | − |E | ≥ 6 ⇒ |E | ≤ 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3|E | ≤ 3|V | − 6
Proof.
sum of region degrees: 2|E |degree of a region ≥ 3⇒ 2|E | ≥ 3|R| ⇒ |R| ≤ 2
3 |E ||V | − |E |+ |R| = 2⇒ |V | − |E |+ 2
3 |E | ≥ 2 ⇒ |V | − 13 |E | ≥ 2
⇒ 3|V | − |E | ≥ 6 ⇒ |E | ≤ 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3|E | ≤ 3|V | − 6
Proof.
sum of region degrees: 2|E |degree of a region ≥ 3⇒ 2|E | ≥ 3|R| ⇒ |R| ≤ 2
3 |E ||V | − |E |+ |R| = 2⇒ |V | − |E |+ 2
3 |E | ≥ 2 ⇒ |V | − 13 |E | ≥ 2
⇒ 3|V | − |E | ≥ 6 ⇒ |E | ≤ 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3|E | ≤ 3|V | − 6
Proof.
sum of region degrees: 2|E |degree of a region ≥ 3⇒ 2|E | ≥ 3|R| ⇒ |R| ≤ 2
3 |E ||V | − |E |+ |R| = 2⇒ |V | − |E |+ 2
3 |E | ≥ 2 ⇒ |V | − 13 |E | ≥ 2
⇒ 3|V | − |E | ≥ 6 ⇒ |E | ≤ 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3:∃v ∈ V [dv ≤ 5]
Proof.
assume: ∀v ∈ V [dv ≥ 6]⇒ 2|E | ≥ 6|V |⇒ |E | ≥ 3|V |⇒ |E | > 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3:∃v ∈ V [dv ≤ 5]
Proof.
assume: ∀v ∈ V [dv ≥ 6]⇒ 2|E | ≥ 6|V |⇒ |E | ≥ 3|V |⇒ |E | > 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3:∃v ∈ V [dv ≤ 5]
Proof.
assume: ∀v ∈ V [dv ≥ 6]⇒ 2|E | ≥ 6|V |⇒ |E | ≥ 3|V |⇒ |E | > 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3:∃v ∈ V [dv ≤ 5]
Proof.
assume: ∀v ∈ V [dv ≥ 6]⇒ 2|E | ≥ 6|V |⇒ |E | ≥ 3|V |⇒ |E | > 3|V | − 6
Planar Graph Theorems
Theorem
G = (V ,E ): connected, planar graph where |V | ≥ 3:∃v ∈ V [dv ≤ 5]
Proof.
assume: ∀v ∈ V [dv ≥ 6]⇒ 2|E | ≥ 6|V |⇒ |E | ≥ 3|V |⇒ |E | > 3|V | − 6
Nonplanar Graphs
Theorem
K5 is not planar.
Proof.
|V | = 5
3|V | − 6 = 3 · 5− 6 = 9
|E | ≤ 9 should hold
but |E | = 10
Nonplanar Graphs
Theorem
K5 is not planar.
Proof.
|V | = 5
3|V | − 6 = 3 · 5− 6 = 9
|E | ≤ 9 should hold
but |E | = 10
Nonplanar Graphs
Theorem
K5 is not planar.
Proof.
|V | = 5
3|V | − 6 = 3 · 5− 6 = 9
|E | ≤ 9 should hold
but |E | = 10
Nonplanar Graphs
Theorem
K5 is not planar.
Proof.
|V | = 5
3|V | − 6 = 3 · 5− 6 = 9
|E | ≤ 9 should hold
but |E | = 10
Nonplanar Graphs
Theorem
K5 is not planar.
Proof.
|V | = 5
3|V | − 6 = 3 · 5− 6 = 9
|E | ≤ 9 should hold
but |E | = 10
Nonplanar Graphs
Theorem
K3,3 is not planar. Proof.
|V | = 6, |E | = 9
if planar then |R| = 5
degree of a region ≥ 4⇒
∑r∈R dr ≥ 20
|E | ≥ 10 should hold
but |E | = 9
Nonplanar Graphs
Theorem
K3,3 is not planar. Proof.
|V | = 6, |E | = 9
if planar then |R| = 5
degree of a region ≥ 4⇒
∑r∈R dr ≥ 20
|E | ≥ 10 should hold
but |E | = 9
Nonplanar Graphs
Theorem
K3,3 is not planar. Proof.
|V | = 6, |E | = 9
if planar then |R| = 5
degree of a region ≥ 4⇒
∑r∈R dr ≥ 20
|E | ≥ 10 should hold
but |E | = 9
Nonplanar Graphs
Theorem
K3,3 is not planar. Proof.
|V | = 6, |E | = 9
if planar then |R| = 5
degree of a region ≥ 4⇒
∑r∈R dr ≥ 20
|E | ≥ 10 should hold
but |E | = 9
Nonplanar Graphs
Theorem
K3,3 is not planar. Proof.
|V | = 6, |E | = 9
if planar then |R| = 5
degree of a region ≥ 4⇒
∑r∈R dr ≥ 20
|E | ≥ 10 should hold
but |E | = 9
Nonplanar Graphs
Theorem
K3,3 is not planar. Proof.
|V | = 6, |E | = 9
if planar then |R| = 5
degree of a region ≥ 4⇒
∑r∈R dr ≥ 20
|E | ≥ 10 should hold
but |E | = 9
Kuratowski’s Theorem
Theorem
G contains a subgraph homeomorphic to K5 or K3,3.⇔
G is not planar.
Platonic Solids
regular polyhedron: a 3-dimensional solidwhere faces are identical regular polygons
projection of a regular polyhedron onto the plane:a planar graph
corners: nodes
sides: edges
faces: regions
Platonic Solid Example
Platonic Solids
|V |: number of corners (nodes)
|E |: number of sides (edges)
|R|: number of faces (regions)
n: number of faces meeting at a corner (node degree)
m: number of sides of a face (region degree)
m, n ≥ 3
2|E | = n · |V |2|E | = m · |R|
Platonic Solids
|V |: number of corners (nodes)
|E |: number of sides (edges)
|R|: number of faces (regions)
n: number of faces meeting at a corner (node degree)
m: number of sides of a face (region degree)
m, n ≥ 3
2|E | = n · |V |2|E | = m · |R|
Platonic Solids
from Euler’s formula:
2 = |V |−|E |+|R| = 2|E |n−|E |+2|E |
m= |E |
(2m −mn + 2n
mn
)> 0
|E |,m, n > 0:
2m −mn + 2n > 0⇒ mn − 2m − 2n < 0
⇒ mn − 2m − 2n + 4 < 4⇒ (m − 2)(n − 2) < 4
only 5 solutions
Platonic Solids
from Euler’s formula:
2 = |V |−|E |+|R| = 2|E |n−|E |+2|E |
m= |E |
(2m −mn + 2n
mn
)> 0
|E |,m, n > 0:
2m −mn + 2n > 0⇒ mn − 2m − 2n < 0
⇒ mn − 2m − 2n + 4 < 4⇒ (m − 2)(n − 2) < 4
only 5 solutions
Platonic Solids
from Euler’s formula:
2 = |V |−|E |+|R| = 2|E |n−|E |+2|E |
m= |E |
(2m −mn + 2n
mn
)> 0
|E |,m, n > 0:
2m −mn + 2n > 0⇒ mn − 2m − 2n < 0
⇒ mn − 2m − 2n + 4 < 4⇒ (m − 2)(n − 2) < 4
only 5 solutions
Platonic Solids
from Euler’s formula:
2 = |V |−|E |+|R| = 2|E |n−|E |+2|E |
m= |E |
(2m −mn + 2n
mn
)> 0
|E |,m, n > 0:
2m −mn + 2n > 0⇒ mn − 2m − 2n < 0
⇒ mn − 2m − 2n + 4 < 4⇒ (m − 2)(n − 2) < 4
only 5 solutions
Platonic Solids
from Euler’s formula:
2 = |V |−|E |+|R| = 2|E |n−|E |+2|E |
m= |E |
(2m −mn + 2n
mn
)> 0
|E |,m, n > 0:
2m −mn + 2n > 0⇒ mn − 2m − 2n < 0
⇒ mn − 2m − 2n + 4 < 4⇒ (m − 2)(n − 2) < 4
only 5 solutions
Tetrahedron
m = 3, n = 3
Hexahedron
m = 4, n = 3
Octahedron
m = 3, n = 4
Dodecahedron
m = 5, n = 3
Icosahedron
m = 3, n = 5
Topics
1 GraphsIntroductionWalksTraversable GraphsPlanar Graphs
2 Graph ProblemsConnectivityGraph ColoringShortest PathTSPSearching Graphs
Connectivity Matrix
A: adjacency matrix of G = (V ,E )
Akij : number of walks of length k between vi and vj
maximum distance between two nodes: |V | − 1
connectivity matrix:C = A1 + A2 + A3 + · · ·+ A|V |−1
connected: all elements of C are non-zero
Connectivity Matrix
A: adjacency matrix of G = (V ,E )
Akij : number of walks of length k between vi and vj
maximum distance between two nodes: |V | − 1
connectivity matrix:C = A1 + A2 + A3 + · · ·+ A|V |−1
connected: all elements of C are non-zero
Warshall’s Algorithm
very expensive to compute the connectivity matrix
easier to find whether there is a walk between two nodesrather than finding the number of walks
for each node:
from all nodes which can reach the current node(rows that contain 1 in current column)
to all nodes which can be reached from the current node(columns that contain 1 in current row)
Warshall’s Algorithm
very expensive to compute the connectivity matrix
easier to find whether there is a walk between two nodesrather than finding the number of walks
for each node:
from all nodes which can reach the current node(rows that contain 1 in current column)
to all nodes which can be reached from the current node(columns that contain 1 in current row)
Warshall’s Algorithm Example
a b c d
a 0 1 0 0b 0 1 0 0c 0 0 0 1d 1 0 1 0
Warshall’s Algorithm Example
a b c d
a 0 1 0 0b 0 1 0 0c 0 0 0 1d 1 1 1 0
Warshall’s Algorithm Example
a b c d
a 0 1 0 0b 0 1 0 0c 0 0 0 1d 1 1 1 0
Warshall’s Algorithm Example
a b c d
a 0 1 0 0b 0 1 0 0c 0 0 0 1d 1 1 1 1
Warshall’s Algorithm Example
a b c d
a 0 1 0 0b 0 1 0 0c 1 1 1 1d 1 1 1 1
Topics
1 GraphsIntroductionWalksTraversable GraphsPlanar Graphs
2 Graph ProblemsConnectivityGraph ColoringShortest PathTSPSearching Graphs
Graph Coloring
G = (V ,E ), C : set of colors
proper coloring of G : find an f : V → C , such that∀(vi , vj) ∈ E [f (vi ) 6= f (vj)]
chromatic number of G : χ(G )minimum |C |finding χ(G ) is a very difficult problem
χ(Kn) = n
Graph Coloring
G = (V ,E ), C : set of colors
proper coloring of G : find an f : V → C , such that∀(vi , vj) ∈ E [f (vi ) 6= f (vj)]
chromatic number of G : χ(G )minimum |C |finding χ(G ) is a very difficult problem
χ(Kn) = n
Graph Coloring
G = (V ,E ), C : set of colors
proper coloring of G : find an f : V → C , such that∀(vi , vj) ∈ E [f (vi ) 6= f (vj)]
chromatic number of G : χ(G )minimum |C |finding χ(G ) is a very difficult problem
χ(Kn) = n
Chromatic Number Example
Herschel graph: χ(G ) = 2
Graph Coloring Solution
pick a node and assign a color
assign same color to all nodes with no conflict
pick an uncolored node and assign a second color
assign same color to all uncolored nodes with no conflict
pick an uncolored node and assign a third color
. . .
Heuristic Solutions
heuristic solution: based on intuition
greedy solution: doesn’t look ahead
doesn’t produce optimal results
Graph Coloring Example
a company produces chemical compounds
some compounds cannot be stored together
such compounds must be placed in separate storage areas
store compounds using minimum number of storage areas
Graph Coloring Example
a company produces chemical compounds
some compounds cannot be stored together
such compounds must be placed in separate storage areas
store compounds using minimum number of storage areas
Graph Coloring Example
every compound is a node
two compounds that cannot be stored together are adjacent
Graph Coloring Example
Graph Coloring Example
Graph Coloring Example
Graph Coloring Example
Graph Coloring Example: Sudoku
every cell is a node
cells of the same roware adjacent
cells of the same columnare adjacent
cells of the same 3× 3 blockare adjacent
every number is a color
problem: properly color a graphthat is partially colored
Graph Coloring Example: Sudoku
every cell is a node
cells of the same roware adjacent
cells of the same columnare adjacent
cells of the same 3× 3 blockare adjacent
every number is a color
problem: properly color a graphthat is partially colored
Region Coloring
coloring a map by assigning different colors to adjacent regions
Theorem (Four Color Theorem)
The regions in a map can be colored using four colors.
Topics
1 GraphsIntroductionWalksTraversable GraphsPlanar Graphs
2 Graph ProblemsConnectivityGraph ColoringShortest PathTSPSearching Graphs
Shortest Path
finding shortest paths from a starting nodeto all other nodes: Dijkstra’s algorithm
Dijkstra’s Algorithm Example
starting node: c
a (∞,−)
b (∞,−)
c (0,−)
f (∞,−)
g (∞,−)
h (∞,−)
Dijkstra’s Algorithm Example
from c: base distance=0
c → f : 6, 6 <∞c → h : 11, 11 <∞
a (∞,−)
b (∞,−)
c (0,−)√
f (6, cf )
g (∞,−)
h (11, ch)
closest node: f
Dijkstra’s Algorithm Example
from c: base distance=0
c → f : 6, 6 <∞c → h : 11, 11 <∞
a (∞,−)
b (∞,−)
c (0,−)√
f (6, cf )
g (∞,−)
h (11, ch)
closest node: f
Dijkstra’s Algorithm Example
from c: base distance=0
c → f : 6, 6 <∞c → h : 11, 11 <∞
a (∞,−)
b (∞,−)
c (0,−)√
f (6, cf )
g (∞,−)
h (11, ch)
closest node: f
Dijkstra’s Algorithm Example
from f : base distance=6
f → a : 6 + 11, 17 <∞f → g : 6 + 9, 15 <∞f → h : 6 + 4, 10 < 11
a (17, cfa)
b (∞,−)
c (0,−)√
f (6, cf )√
g (15, cfg)
h (10, cfh)
closest node: h
Dijkstra’s Algorithm Example
from f : base distance=6
f → a : 6 + 11, 17 <∞f → g : 6 + 9, 15 <∞f → h : 6 + 4, 10 < 11
a (17, cfa)
b (∞,−)
c (0,−)√
f (6, cf )√
g (15, cfg)
h (10, cfh)
closest node: h
Dijkstra’s Algorithm Example
from f : base distance=6
f → a : 6 + 11, 17 <∞f → g : 6 + 9, 15 <∞f → h : 6 + 4, 10 < 11
a (17, cfa)
b (∞,−)
c (0,−)√
f (6, cf )√
g (15, cfg)
h (10, cfh)
closest node: h
Dijkstra’s Algorithm Example
from h: base distance=10
h→ a : 10 + 11, 21 ≮ 17
h→ g : 10 + 4, 14 < 15
a (17, cfa)
b (∞,−)
c (0,−)√
f (6, cf )√
g (14, cfhg)
h (10, cfh)√
closest node: g
Dijkstra’s Algorithm Example
from h: base distance=10
h→ a : 10 + 11, 21 ≮ 17
h→ g : 10 + 4, 14 < 15
a (17, cfa)
b (∞,−)
c (0,−)√
f (6, cf )√
g (14, cfhg)
h (10, cfh)√
closest node: g
Dijkstra’s Algorithm Example
from h: base distance=10
h→ a : 10 + 11, 21 ≮ 17
h→ g : 10 + 4, 14 < 15
a (17, cfa)
b (∞,−)
c (0,−)√
f (6, cf )√
g (14, cfhg)
h (10, cfh)√
closest node: g
Dijkstra’s Algorithm Example
from g : base distance=14
g → a : 14 + 17, 31 ≮ 17
a (17, cfa)
b (∞,−)
c (0,−)√
f (6, cf )√
g (14, cfhg)√
h (10, cfh)√
closest node: a
Dijkstra’s Algorithm Example
from g : base distance=14
g → a : 14 + 17, 31 ≮ 17
a (17, cfa)
b (∞,−)
c (0,−)√
f (6, cf )√
g (14, cfhg)√
h (10, cfh)√
closest node: a
Dijkstra’s Algorithm Example
from g : base distance=14
g → a : 14 + 17, 31 ≮ 17
a (17, cfa)
b (∞,−)
c (0,−)√
f (6, cf )√
g (14, cfhg)√
h (10, cfh)√
closest node: a
Dijkstra’s Algorithm Example
from a: base distance=17
a→ b : 17 + 5, 22 <∞a (17, cfa)
√
b (22, cfab)
c (0,−)√
f (6, cf )√
g (14, cfhg)√
h (10, cfh)√
last node: b
Dijkstra’s Algorithm Example
from a: base distance=17
a→ b : 17 + 5, 22 <∞a (17, cfa)
√
b (22, cfab)
c (0,−)√
f (6, cf )√
g (14, cfhg)√
h (10, cfh)√
last node: b
Dijkstra’s Algorithm Example
from a: base distance=17
a→ b : 17 + 5, 22 <∞a (17, cfa)
√
b (22, cfab)
c (0,−)√
f (6, cf )√
g (14, cfhg)√
h (10, cfh)√
last node: b
Topics
1 GraphsIntroductionWalksTraversable GraphsPlanar Graphs
2 Graph ProblemsConnectivityGraph ColoringShortest PathTSPSearching Graphs
Traveling Salesperson Problem
start from a home town
visit every city exactly once
return to the home town
minimum total distance
find Hamiltonian cycle
very difficult problem
Traveling Salesperson Problem
start from a home town
visit every city exactly once
return to the home town
minimum total distance
find Hamiltonian cycle
very difficult problem
TSP Solution
heuristic: nearest-neighbor
Topics
1 GraphsIntroductionWalksTraversable GraphsPlanar Graphs
2 Graph ProblemsConnectivityGraph ColoringShortest PathTSPSearching Graphs
Searching Graphs
searching nodes of graph G = (V ,E ) starting from node v1
depth-first
breadth-first
Depth-First Search
1 v ← v1,T = ∅, D = {v1}2 find smallest i in 2 ≤ i ≤ |V | such that (v , vi ) ∈ E and vi /∈ D
if no such i : go to step 3if found: T = T ∪ {(v , vi )}, D = D ∪ {vi}, v ← vi ,go to step 2
3 if v = v1: result is T
4 if v 6= v1: v ← backtrack(v), go to step 2
Depth-First Search
1 v ← v1,T = ∅, D = {v1}2 find smallest i in 2 ≤ i ≤ |V | such that (v , vi ) ∈ E and vi /∈ D
if no such i : go to step 3if found: T = T ∪ {(v , vi )}, D = D ∪ {vi}, v ← vi ,go to step 2
3 if v = v1: result is T
4 if v 6= v1: v ← backtrack(v), go to step 2
Depth-First Search
1 v ← v1,T = ∅, D = {v1}2 find smallest i in 2 ≤ i ≤ |V | such that (v , vi ) ∈ E and vi /∈ D
if no such i : go to step 3if found: T = T ∪ {(v , vi )}, D = D ∪ {vi}, v ← vi ,go to step 2
3 if v = v1: result is T
4 if v 6= v1: v ← backtrack(v), go to step 2
Depth-First Search
1 v ← v1,T = ∅, D = {v1}2 find smallest i in 2 ≤ i ≤ |V | such that (v , vi ) ∈ E and vi /∈ D
if no such i : go to step 3if found: T = T ∪ {(v , vi )}, D = D ∪ {vi}, v ← vi ,go to step 2
3 if v = v1: result is T
4 if v 6= v1: v ← backtrack(v), go to step 2
Breadth-First Search
1 T = ∅, D = {v1}, Q = (v1)
2 if Q empty: result is T
3 if Q not empty: v ← front(Q), Q ← Q − vfor 2 ≤ i ≤ |V | check edges (v , vi ) ∈ E :
if vi /∈ D : Q = Q + vi , T = T ∪ {(v , vi )}, D = D ∪ {vi}go to step 3
Breadth-First Search
1 T = ∅, D = {v1}, Q = (v1)
2 if Q empty: result is T
3 if Q not empty: v ← front(Q), Q ← Q − vfor 2 ≤ i ≤ |V | check edges (v , vi ) ∈ E :
if vi /∈ D : Q = Q + vi , T = T ∪ {(v , vi )}, D = D ∪ {vi}go to step 3
References
Required Reading: Grimaldi
Chapter 11: An Introduction to Graph Theory
Chapter 7: Relations: The Second Time Around
7.2. Computer Recognition: Zero-One Matricesand Directed Graphs
Chapter 13: Optimization and Matching
13.1. Dijkstra’s Shortest Path Algorithm
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